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2 years ago

Active peer pressure involves all of the following EXCEPT: A. bribery B. teasing C. put-downs D. famous people

Mathematics

2 answers:

lana [24]2 years ago

7 0

Answer:

Step-by-step explanation:

it is d because famous people might do it but it is the lest likey

Morgarella [4.7K]2 years ago

7 0

Answer:C

Step-by-step explanation:

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the waverly brush company issued 4,000 sharesof common sotck worth $200,000.00 total. what is the par value of each share?

erica [24]

The par value of each share is $50. x = 200,000/4,000 x = 50

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2 years ago

In a sample of 10 cards, 4 are red and 6 are blue. If 2 cards are selected at random from the sample, one at a time without repl

Charra [1.4K]

The exact answer in fraction from is 2/15

The approximate answer in decimal form is 0.1333 (rounded to four decimal places)

The decimal value 0.1333 is equivalent to 13.33%

These are three ways to say the same basic answer.

===============================================

Explanation:

We have 4 red cards and 6 blue cards giving 4+6 = 10 total.

The probability of picking not blue (aka picking red) is 4/10 because there are 4 red out of 10 total.

If we don't replace the card we picked, then we have 3 red left over out of 9 total. The probability of picking red again is 3/9

Multiply those fractions
(4/10)*(3/9) = (4*3)/(10*9) = 12/90

Now reduce
12/90 = (6*2)/(6*15) = 2/15
note how the GCF 6 cancels

Using a calculator,
2/15 = 0.1333 approximately
The '3's after the decimal point go on forever.

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2 years ago

The value of two numbers if their sum is 24 and their difference is 4.

ankoles [38]

Answer:

14 and 10

Step-by-step explanation:

14+10=24

14-10=4

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2 years ago

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hit me up on Instagram

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2 years ago

Read 2 more answers

for a class project, a political science student at a large university wants to estimate the percent of students that are regist

nekit [7.7K]

Answer:

90% confidence interval for the true percent of students that are registered voters is [0.56 , 0.64].

Step-by-step explanation:

We are given that for a class project, a political science student at a large university wants to estimate the percent of students that are registered voters.

He surveys 500 students and finds that 300 are registered voters.

Firstly, the pivotal quantity for 90% confidence interval for the true proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of students who are registered voters = $\frac{300}{500}$ = 0.60

n = sample of students = 500

p = true percent of students

Here for constructing 90% confidence interval we have used One-sample z proportion test statistics.

So, 90% confidence interval for the true proportion, p is ;

P(-1.6449 < N(0,1) < 1.6449) = 0.90 {As the critical value of z at 5%

level of significance are -1.6449 & 1.6449}

P(-1.6449 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.6449) = 0.90

P( $-1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

90% confidence interval for p =[ $\hat p-1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.60-1.6449 \times {\sqrt{\frac{0.60(1-0.60)}{500} } }$ , $0.60+1.6449 \times {\sqrt{\frac{0.60(1-0.60)}{500} } }$ ]

= [0.56 , 0.64]

Therefore, 90% confidence interval for the true percent of students that are registered voters is [0.56 , 0.64].

Interpretation of the above confidence interval is that we are 90% confident that the true percent of students that are registered voters will lie between 0.56 and 0.64.

6 0

2 years ago